A student calculates the density of an unknown metal sample to be 8.25 g/cm3

IrinaVladis [17]

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5 0

3 years ago

The rate constant for a certain reaction is k = 4.70×10−3 s−1 . If the initial reactant concentration was 0.700 M, what will the

Lynna [10]

Answer:

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

Explanation:

The unit of k is s⁻¹.

The order of the reaction = first order.

First order reaction: A first order reaction is a reaction in which the rate of reaction depends only the value of the concentration of the reactant.

$-\frac{d[A]}{dt} =kt$

[A] = the concentration of the reactant at time t

k= rate constant

t= time

Here k= 4.70×10⁻³ s⁻¹

t= 4.00

[A₀] = initial concentration of reactant = 0.700 M

$-\frac{d[A]}{dt} =kt$

$\Rightarrow -\frac{d[A]}{[A]}=kdt$

Integrating both sides

$\Rightarrow\int -\frac{d[A]}{[A]}=\int kdt$

⇒ -ln[A] = kt +c

When t=0 , [A] =[A₀]

-ln[A₀] = k.0 + c

⇒c= -ln[A₀]

Therefore

-ln[A] = kt - ln[A₀]

Putting the value of k, [A₀] and t

- ln[A] =4.70×10⁻³×4 -ln (0.70)

⇒-ln[A]= 0.375

⇒[A] = 0.686

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

5 0

3 years ago

Which sample of matter is classified as a solution? 1. H2O(s) 2. H2O(l) 3. CO2(g) 4. CO2(aq)

lys-0071 [83]

Answer : Option 4) $CO_{2}_{(aq)}$

Explanation : $CO_{2}_{(aq)}$ is the only sample of matter which can be classified as a solution. As the solution can be defined as a liquid mixture which contains a minor component (the solute) that is uniformly distributed within the major component (the solvent).

In this case, the solute is $CO_{2}$ which is dissolved in water which acts as an solvent. Also, it has a subscript which is aq. which means aqueous, is often given to the solution in which the solvent is water.

Therefore, $CO_{2}_{(aq)}$ is the correct answer.

4 0

3 years ago

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A 2.026g sample of a hydrate of sodium carbonate (Na2CO3) is heated to remove water. After heating, the mass of the sample is 0.

OLEGan [10]

Answer:

62.98 % of the sample of hydrate is water

Explanation:

Step 1: Data given

Mass of the sample of a hydrate of sodium carbonate (Na2CO3) = 2.026 grams

After heating, the mass of the sample is 0.750 g

Molar mass H2O = 18.02 g/mol

Step 2: Calculate mass of water

Mass water = mass of hydrate - mass of sample after heating

Mass water = 2.026 grams - 0.750 grams

Mass water = 1.276 grams

Step 3: Calculate mass % percent of water

Mass % of water = (mass of water / total mass hydrate) * 100 %

Mass % of water = (1.276 grams / 2.026 grams) *100 %

Mass % of water = 62.98 %

62.98 % of the sample of hydrate is water

7 0

3 years ago

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A solution contains [Ba2+] = 5.0 × 10−5 M, [Zn2+] = 2.0 × 10−7 M, and [Ag+] = 3.0 × 10−5 M. Sodium oxalate (Na2C2O4) is slowly a

Jet001 [13]

Answer:

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

Explanation:

1. Calculate the equilibrium concentrations of oxalate ion

Let [C₂O₄²⁻] = c

(a) Barium oxalate

BaC₂O₄ ⇌ Ba²⁺ + C₂O₄²⁻

E/mol·L⁻¹: 5.0 × 10⁻⁵ c

Ksp = [Ba²⁺][C₂O₄²⁻] = 5.0 × 10⁻⁵c = 1.5 × 10⁻⁸

c = (1.5 × 10⁻⁸)/(5.0 × 10⁻⁵) = 3.0 × 10⁻⁴ mol·L⁻¹

(b) Zinc oxalate

ZnC₂O₄ ⇌ Zn²⁺ + C₂O₄²⁻

E/mol·L⁻¹: 2.0 × 10⁻⁷ c

Ksp = [Zn²⁺][C₂O₄²⁻] = 2.0 × 10⁻⁷c = 1.35 × 10⁻⁹

c = (1.35 × 10⁻⁹)/(2.0 × 10⁻⁷) = 6.8 × 10⁻³ mol·L⁻¹

(c) Silver oxalate

Ag₂C₂O₄ ⇌ 2Ag⁺ + C₂O₄²⁻

E/mol·L⁻¹: 3.0 × 10⁻⁵ c

Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹

c = (1.1 × 10⁻¹¹)/(9.0 × 10⁻¹⁰) = 0.012 mol·L⁻¹

2. Decide the order of precipitation

BaC₂O₄ will precipitate when c > 3.0 × 10⁻⁴ mol·L⁻¹

ZnC₂O₄ will precipitate when c > 6.8 × 10⁻³ mol·L⁻¹

Ag₂C₂O₄ will precipitate when c > 0.028 mol·L⁻¹

This happens to be the order of increasing concentration of oxalate ion.

The order of precipitation is

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

4 0

3 years ago